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Number of turns to reach from one node to other in binary tree
  • Difficulty Level : Hard
  • Last Updated : 21 Apr, 2021

Given a binary tree and two nodes. The task is to count the number of turns needs to reach from one node to another node of the Binary tree.
Examples: 
 

Input:   Below Binary Tree and two nodes
        5 & 6 
                   1
                /     \
               2        3
             /   \    /   \
            4     5   6     7
           /         / \
          8         9   10
Output: Number of Turns needed to reach 
from 5 to 6:  3
        
Input: For above tree if two nodes are 1 & 4
Output: Straight line : 0 turn 

 

Idea based on the Lowest Common Ancestor in a Binary Tree 
We have to follow the step. 
1…Find the LCA of given two node 
2…Given node present either on the left side or right side or equal to LCA. 
…. According to above condition over program falls under two Case. 
 

Case 1:  
If none of the nodes is equal to
LCA, we get these nodes either on
the left side or right side. 
We call two functions for each node.
....a)  if (CountTurn(LCA->right, first, 
                          false, &Count)
        || CountTurn(LCA->left, first, 
                          true, &Count)) ; 
....b)  Same for second node.
....Here Count is used to store number of 
turns need to reached the target node.    

Case 2:  
If one of the nodes is equal to LCA_Node. 
Then we count only number of turns needs
to reached the second node.
If LCA == (Either first or second)
....a)  if (countTurn(LCA->right, second/first, 
                                false, &Count)  
         || countTurn(LCA->left, second/first, 
                              true, &Count)) ; 

3… Working of CountTurn Function 
// we pass turn true if we move 
// left subtree and false if we 
// move right subTree 
 

CountTurn(LCA, Target_node, count, Turn)

// if found the key value in tree 
if (root->key == key)
    return true;
case 1: 
   If Turn is true that means we are 
      in left_subtree  
   If we going left_subtree then there 
      is no need to increment count 
   else
      Increment count and set turn as false  
case 2:
   if Turn is false that means we are in
      right_subtree    
   if we going right_subtree then there is
      no need to increment count else
   increment count and set turn as true.

// if key is not found.
return false;
 

Below the implementation of above idea. 
 



C++




// C++ Program to count number of turns
// in a Binary Tree.
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    struct Node* left, *right;
    int key;
};
 
// Utility function to create a new
// tree Node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
// Utility function to find the LCA of
// two given values n1 and n2.
struct Node* findLCA(struct Node* root,
                        int n1, int n2)
{
    // Base case
    if (root == NULL)
        return NULL;
 
    // If either n1 or n2 matches with
    // root's key, report the presence by
    // returning root (Note that if a key
    // is ancestor of other, then the
    // ancestor key becomes LCA
    if (root->key == n1 || root->key == n2)
        return root;
 
    // Look for keys in left and right subtrees
    Node* left_lca = findLCA(root->left, n1, n2);
    Node* right_lca = findLCA(root->right, n1, n2);
 
    // If both of the above calls return
    // Non-NULL, then one key is present
    // in once subtree and other is present
    // in other, So this node is the LCA
    if (left_lca && right_lca)
        return root;
 
    // Otherwise check if left subtree or right
    // subtree is LCA
    return (left_lca != NULL) ? left_lca :
                                right_lca;
}
 
// function count number of turn need to reach
// given node from it's LCA we have two way to
bool CountTurn(Node* root, int key, bool turn,
                                   int* count)
{
    if (root == NULL)
        return false;
 
    // if found the key value in tree
    if (root->key == key)
        return true;
 
    // Case 1:
    if (turn == true) {
        if (CountTurn(root->left, key, turn, count))
            return true;
        if (CountTurn(root->right, key, !turn, count)) {
            *count += 1;
            return true;
        }
    }
    else // Case 2:
    {
        if (CountTurn(root->right, key, turn, count))
            return true;
        if (CountTurn(root->left, key, !turn, count)) {
            *count += 1;
            return true;
        }
    }
    return false;
}
 
// Function to find nodes common to given two nodes
int NumberOFTurn(struct Node* root, int first,
                                   int second)
{
    struct Node* LCA = findLCA(root, first, second);
 
    // there is no path between these two node
    if (LCA == NULL)
        return -1;
    int Count = 0;
 
    // case 1:
    if (LCA->key != first && LCA->key != second) {
         
        // count number of turns needs to reached
        // the second node from LCA
        if (CountTurn(LCA->right, second, false,
                                           &Count)
            || CountTurn(LCA->left, second, true,
                                           &Count))
            ;
         
        // count number of turns needs to reached
        // the first node from LCA
        if (CountTurn(LCA->left, first, true,
                                       &Count)
            || CountTurn(LCA->right, first, false,
                                       &Count))
            ;
        return Count + 1;
    }
 
    // case 2:
    if (LCA->key == first) {
 
        // count number of turns needs to reached
        // the second node from LCA
        CountTurn(LCA->right, second, false, &Count);
        CountTurn(LCA->left, second, true, &Count);
        return Count;
    } else {
 
        // count number of turns needs to reached
        // the first node from LCA1
        CountTurn(LCA->right, first, false, &Count);
        CountTurn(LCA->left, first, true, &Count);
        return Count;
    }
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree given in the above
    // example
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->right->left->left = newNode(9);
    root->right->left->right = newNode(10);
 
    int turn = 0;
    if ((turn = NumberOFTurn(root, 5, 10)))
        cout << turn << endl;
    else
        cout << "Not Possible" << endl;
 
    return 0;
}

Java




//A Java Program to count number of turns
//in a Binary Tree.
public class Turns_to_reach_another_node {
 
    // making Count global such that it can get
    // modified by different methods
    static int Count;
 
    // A Binary Tree Node
    static class Node {
        Node left, right;
        int key;
 
        // Constructor
        Node(int key) {
            this.key = key;
            left = null;
            right = null;
        }
    }
 
    // Utility function to find the LCA of
    // two given values n1 and n2.
    static Node findLCA(Node root, int n1, int n2) {
        // Base case
        if (root == null)
            return null;
 
        // If either n1 or n2 matches with
        // root's key, report the presence by
        // returning root (Note that if a key
        // is ancestor of other, then the
        // ancestor key becomes LCA
        if (root.key == n1 || root.key == n2)
            return root;
 
        // Look for keys in left and right subtrees
        Node left_lca = findLCA(root.left, n1, n2);
        Node right_lca = findLCA(root.right, n1, n2);
 
        // If both of the above calls return
        // Non-NULL, then one key is present
        // in once subtree and other is present
        // in other, So this node is the LCA
        if (left_lca != null && right_lca != null)
            return root;
 
        // Otherwise check if left subtree or right
        // subtree is LCA
        return (left_lca != null) ? left_lca : right_lca;
    }
 
    // function count number of turn need to reach
    // given node from it's LCA we have two way to
    static boolean CountTurn(Node root, int key, boolean turn) {
        if (root == null)
            return false;
 
        // if found the key value in tree
        if (root.key == key)
            return true;
 
        // Case 1:
        if (turn == true) {
            if (CountTurn(root.left, key, turn))
                return true;
            if (CountTurn(root.right, key, !turn)) {
                Count += 1;
                return true;
            }
        } else // Case 2:
        {
            if (CountTurn(root.right, key, turn))
                return true;
            if (CountTurn(root.left, key, !turn)) {
                Count += 1;
                return true;
            }
        }
        return false;
    }
 
    // Function to find nodes common to given two nodes
    static int NumberOfTurn(Node root, int first, int second) {
        Node LCA = findLCA(root, first, second);
 
        // there is no path between these two node
        if (LCA == null)
            return -1;
        Count = 0;
 
        // case 1:
        if (LCA.key != first && LCA.key != second) {
 
            // count number of turns needs to reached
            // the second node from LCA
            if (CountTurn(LCA.right, second, false)
                    || CountTurn(LCA.left, second, true))
                ;
 
            // count number of turns needs to reached
            // the first node from LCA
            if (CountTurn(LCA.left, first, true)
                    || CountTurn(LCA.right, first, false))
                ;
            return Count + 1;
        }
 
        // case 2:
        if (LCA.key == first) {
 
            // count number of turns needs to reached
            // the second node from LCA
            CountTurn(LCA.right, second, false);
            CountTurn(LCA.left, second, true);
            return Count;
        } else {
 
            // count number of turns needs to reached
            // the first node from LCA1
            CountTurn(LCA.right, first, false);
            CountTurn(LCA.left, first, true);
            return Count;
        }
    }
 
    // Driver program to test above functions
    public static void main(String[] args) {
        // Let us create binary tree given in the above
        // example
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
        root.left.left.left = new Node(8);
        root.right.left.left = new Node(9);
        root.right.left.right = new Node(10);
 
        int turn = 0;
        if ((turn = NumberOfTurn(root, 5, 10)) != 0)
            System.out.println(turn);
        else
            System.out.println("Not Possible");
    }
 
}
// This code is contributed by Sumit Ghosh

Python3




# Python Program to count number of turns
# in a Binary Tree.
 
# A Binary Tree Node
class Node:
    def __init__(self):
        self.key = 0
        self.left = None
        self.right = None
 
# Utility function to create a new
# tree Node
def newNode(key: int) -> Node:
    temp = Node()
    temp.key = key
    temp.left = None
    temp.right = None
    return temp
 
# Utility function to find the LCA of
# two given values n1 and n2.
def findLCA(root: Node, n1: int, n2: int) -> Node:
 
    # Base case
    if root is None:
        return None
 
    # If either n1 or n2 matches with
    # root's key, report the presence by
    # returning root (Note that if a key
    # is ancestor of other, then the
    # ancestor key becomes LCA
    if root.key == n1 or root.key == n2:
        return root
 
    # Look for keys in left and right subtrees
    left_lca = findLCA(root.left, n1, n2)
    right_lca = findLCA(root.right, n1, n2)
 
    # If both of the above calls return
    # Non-NULL, then one key is present
    # in once subtree and other is present
    # in other, So this node is the LCA
    if left_lca and right_lca:
        return root
 
    # Otherwise check if left subtree or right
    # subtree is LCA
    return (left_lca if left_lca is not None else right_lca)
 
# function count number of turn need to reach
# given node from it's LCA we have two way to
def countTurn(root: Node, key: int, turn: bool) -> bool:
    global count
    if root is None:
        return False
 
    # if found the key value in tree
    if root.key == key:
        return True
 
    # Case 1:
    if turn is True:
        if countTurn(root.left, key, turn):
            return True
        if countTurn(root.right, key, not turn):
            count += 1
            return True
 
    # Case 2:
    else:
        if countTurn(root.right, key, turn):
            return True
        if countTurn(root.left, key, not turn):
            count += 1
            return True
    return False
 
# Function to find nodes common to given two nodes
def numberOfTurn(root: Node, first: int, second: int) -> int:
    global count
    LCA = findLCA(root, first, second)
 
    # there is no path between these two node
    if LCA is None:
        return -1
 
    count = 0
 
    # case 1:
    if LCA.key != first and LCA.key != second:
 
        # count number of turns needs to reached
        # the second node from LCA
        if countTurn(LCA.right, second, False) or countTurn(
                LCA.left, second, True):
            pass
 
        # count number of turns needs to reached
        # the first node from LCA
        if countTurn(LCA.left, first, True) or countTurn(
                LCA.right, first, False):
            pass
        return count + 1
 
    # case 2:
    if LCA.key == first:
 
        # count number of turns needs to reached
        # the second node from LCA
        countTurn(LCA.right, second, False)
        countTurn(LCA.left, second, True)
        return count
    else:
 
        # count number of turns needs to reached
        # the first node from LCA1
        countTurn(LCA.right, first, False)
        countTurn(LCA.left, first, True)
        return count
 
# Driver Code
if __name__ == "__main__":
    count = 0
 
    # Let us create binary tree given in the above
    # example
    root = Node()
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(4)
    root.left.right = newNode(5)
    root.right.left = newNode(6)
    root.right.right = newNode(7)
    root.left.left.left = newNode(8)
    root.right.left.left = newNode(9)
    root.right.left.right = newNode(10)
 
    turn = numberOfTurn(root, 5, 10)
    if turn:
        print(turn)
    else:
        print("Not possible")
 
# This code is contributed by
# sanjeev2552

C#




using System;
 
//A C# Program to count number of turns
//in a Binary Tree.
public class Turns_to_reach_another_node
{
 
    // making Count global such that it can get
    // modified by different methods
    public static int Count;
 
    // A Binary Tree Node
    public class Node
    {
        public Node left, right;
        public int key;
 
        // Constructor
        public Node(int key)
        {
            this.key = key;
            left = null;
            right = null;
        }
    }
 
    // Utility function to find the LCA of
    // two given values n1 and n2.
    public static Node findLCA(Node root, int n1, int n2)
    {
        // Base case
        if (root == null)
        {
            return null;
        }
 
        // If either n1 or n2 matches with
        // root's key, report the presence by
        // returning root (Note that if a key
        // is ancestor of other, then the
        // ancestor key becomes LCA
        if (root.key == n1 || root.key == n2)
        {
            return root;
        }
 
        // Look for keys in left and right subtrees
        Node left_lca = findLCA(root.left, n1, n2);
        Node right_lca = findLCA(root.right, n1, n2);
 
        // If both of the above calls return
        // Non-NULL, then one key is present
        // in once subtree and other is present
        // in other, So this node is the LCA
        if (left_lca != null && right_lca != null)
        {
            return root;
        }
 
        // Otherwise check if left subtree or right
        // subtree is LCA
        return (left_lca != null) ? left_lca : right_lca;
    }
 
    // function count number of turn need to reach
    // given node from it's LCA we have two way to
    public static bool CountTurn(Node root, int key, bool turn)
    {
        if (root == null)
        {
            return false;
        }
 
        // if found the key value in tree
        if (root.key == key)
        {
            return true;
        }
 
        // Case 1:
        if (turn == true)
        {
            if (CountTurn(root.left, key, turn))
            {
                return true;
            }
            if (CountTurn(root.right, key, !turn))
            {
                Count += 1;
                return true;
            }
        }
        else // Case 2:
        {
            if (CountTurn(root.right, key, turn))
            {
                return true;
            }
            if (CountTurn(root.left, key, !turn))
            {
                Count += 1;
                return true;
            }
        }
        return false;
    }
 
    // Function to find nodes common to given two nodes
    public static int NumberOfTurn(Node root, int first, int second)
    {
        Node LCA = findLCA(root, first, second);
 
        // there is no path between these two node
        if (LCA == null)
        {
            return -1;
        }
        Count = 0;
 
        // case 1:
        if (LCA.key != first && LCA.key != second)
        {
 
            // count number of turns needs to reached
            // the second node from LCA
            if (CountTurn(LCA.right, second, false)
                || CountTurn(LCA.left, second, true))
            {
                ;
            }
 
            // count number of turns needs to reached
            // the first node from LCA
            if (CountTurn(LCA.left, first, true)
                || CountTurn(LCA.right, first, false))
            {
                ;
            }
            return Count + 1;
        }
 
        // case 2:
        if (LCA.key == first)
        {
 
            // count number of turns needs to reached
            // the second node from LCA
            CountTurn(LCA.right, second, false);
            CountTurn(LCA.left, second, true);
            return Count;
        }
        else
        {
 
            // count number of turns needs to reached
            // the first node from LCA1
            CountTurn(LCA.right, first, false);
            CountTurn(LCA.left, first, true);
            return Count;
        }
    }
 
    // Driver program to test above functions
    public static void Main(string[] args)
    {
        // Let us create binary tree given in the above
        // example
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.left = new Node(6);
        root.right.right = new Node(7);
        root.left.left.left = new Node(8);
        root.right.left.left = new Node(9);
        root.right.left.right = new Node(10);
 
        int turn = 0;
        if ((turn = NumberOfTurn(root, 5, 10)) != 0)
        {
            Console.WriteLine(turn);
        }
        else
        {
            Console.WriteLine("Not Possible");
        }
    }
 
}
 
// This code is contributed by Shrikant13
Output
4

Time Complexity : O(n)

Alternate Solution:

We have to follow the step.  

  1.  Find the LCA of given two-node
  2.  Store the path of two-node from LCA in strings S1 and S2 that will store ‘l’  if we have to take a left turn in the path starting from LCA to that node and ‘r’ if we take a right turn in the path starting from LCA.
  3.  Reverse one of the strings and concatenate both strings. 
  4. Count number of time characters in our resultant string not equal to its adjacent character.

C++




// C++ Program to count number of turns
// in a Binary Tree.
#include <bits/stdc++.h>
using namespace std;
 
// A Binary Tree Node
struct Node {
    struct Node* left, *right;
    int key;
};
 
// Utility function to create a new
// tree Node
Node* newNode(int key)
{
    Node* temp = new Node;
    temp->key = key;
    temp->left = temp->right = NULL;
    return temp;
}
 
//Preorder traversal to store l or r in the string traversing
//from LCA to the given node key
void findPath(struct Node* root, int d,string str,string &s){
    if(root==NULL){
        return;
    }
     
    if(root->key==d){
        s=str;
        return;
    }
    findPath(root->left,d,str+"l",s);
    findPath(root->right,d,str+"r",s);
}
 
// Utility function to find the LCA of
// two given values n1 and n2.
struct Node* findLCAUtil(struct Node* root, int n1, int n2,bool&v1,bool&v2){
    // Base case
    if (root == NULL)
        return NULL;
 
    if (root->key == n1){
        v1=true;
        return root;
    }
    if(root->key == n2){
        v2=true;
        return root;
    }
    // Look for keys in left and right subtrees
    Node* left_lca = findLCAUtil(root->left, n1, n2,v1,v2);
    Node* right_lca = findLCAUtil(root->right, n1, n2,v1,v2);
    if (left_lca && right_lca){
        return root;
    }
    return (left_lca != NULL) ? left_lca : right_lca;
}
 
bool find(Node* root, int x){
    if(root==NULL){
        return false;
    }
    if((root->key==x) || find(root->left,x) || find(root->right,x)){
        return true;
    }
    return false;
}
 
//Function should return LCA of two nodes if both nodes are
//present otherwise should return NULL
Node* findLCA(struct Node* root, int first, int second){
    bool v1=false;
    bool v2=false;
    Node* lca = findLCAUtil(root,first,second,v1,v2);
    if((v1&&v2) || (v1&&find(lca,second)) || (v2&&find(lca,first))){
        return lca;
    }
    return NULL;
}
   
 
// function should return the number of turns required to go from
//first node to second node
int NumberOFTurns(struct Node* root, int first, int second){
  // base cases if root is not present or both node values are same
    if(root==NULL || (first==second)){
        return 0;
    }
   
    string s1 ="";
    string s2 = "";
    Node* lca = findLCA(root,first,second);
 
    findPath(lca,first,"",s1);
    findPath(lca,second,"",s2);
    if(s1.length()==0 && s2.length()==0){
       return -1;
    }
    reverse(s1.begin(),s1.end());
    s1+=s2;
 
    int cnt=0;
    bool flag=false;
    for(int i=0; i<(s1.length()-1); i++){
        if(s1[i]!=s1[i+1]){
            flag=true;
            cnt+=1;
        }
    }
    if(!flag){
       return -1;
    }
    return cnt;
}
 
// Driver program to test above functions
int main()
{
    // Let us create binary tree given in the above
    // example
    Node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->left = newNode(6);
    root->right->right = newNode(7);
    root->left->left->left = newNode(8);
    root->right->left->left = newNode(9);
    root->right->left->right = newNode(10);
 
    int turn = 0;
    if ((turn = NumberOFTurns(root, 5, 10))!=-1)
        cout << turn << endl;
    else
        cout << "Not Possible" << endl;
 
    return 0;
}
Output
4

Time Complexity: O(n) 

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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